Spring momentum conservation problem

  • #1
member 731016
Homework Statement
I have successfully solved the problem below by assuming that momentum is conserved and that there is an inelastic collision occurring between the masses. However, I am wondering whether the momentum being conserved is a valid assumption that I have made.
Relevant Equations
##E_i = E_f##
##\vec p_i = \vec p_f##
For this problem,
1692395139491.png

The reason why I am not sure whether it is a valid assumption whether momentum is conserved because during the collision if we consider the two masses to be the system, then there will be a uniform gravitational field acting on both masses, and a spring force that is acting upwards. Therefore, there will be two external forces acting on the system. The only reason I can think of for momentum being conserved in this case is if the forces acting on the both the masses acted over such a short time interval that there was no change in the momentum due to the forces.

However, if we define the system as everything, the two masses, the spring, and the source of the g-field, then I believe everything is internal force pairs so momentum is conserved.

If someone please knows whether momentum is conserved is a valid assumption and why, that would be greatly appreciated!

Many thanks!
 
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  • #2
ChiralSuperfields said:
The only reason I can think of for momentum being conserved in this case is if the forces acting on the both the masses acted over such a short time interval that there was no change in the momentum due to the forces.
This reasoning is correct. Over the time interval that the masses stick together they are not displaced appreciably in the gravitational field.
 
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  • #3
ChiralSuperfields said:
Homework Statement: I have successfully solved the problem below by assuming that momentum is conserved and that there is an inelastic collision occurring between the masses. However, I am wondering whether the momentum being conserved is a valid assumption that I have made.
Relevant Equations: ##E_i = E_f##
##\vec p_i = \vec p_f##

For this problem,
View attachment 330735
The reason why I am not sure whether it is a valid assumption whether momentum is conserved because during the collision if we consider the two masses to be the system, then there will be a uniform gravitational field acting on both masses, and a spring force that is acting upwards. Therefore, there will be two external forces acting on the system. The only reason I can think of for momentum being conserved in this case is if the forces acting on the both the masses acted over such a short time interval that there was no change in the momentum due to the forces.

However, if we define the system as everything, the two masses, the spring, and the source of the g-field, then I believe everything is internal force pairs so momentum is conserved.

If someone please knows whether momentum is conserved is a valid assumption and why, that would be greatly appreciated!

Many thanks!
The momentum of the 1.0 kg mass is clearly not conserved. Neither is the momentum of the 2.0 kg mass. But if you take the 1.0 kg mass, the 2.0 kg mass, and the spring (which is ideal, and thus massless) to be your system, then the momentum is conserved, (If you want to be picky, throw the Earth into this and use the Newtonian gravitational potential energy.)

-Dan
 
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  • #4
topsquark said:
The momentum of the 1.0 kg mass is clearly not conserved. Neither is the momentum of the 2.0 kg mass. But if you take the 1.0 kg mass, the 2.0 kg mass, and the spring (which is ideal, and thus massless) to be your system, then the momentum is conserved, (If you want to be picky, throw the Earth into this and use the Newtonian gravitational potential energy.)

-Dan
That doesn’t really help. The issue is the time interval over which momentum is to be considered.
If you take the time up until the dropped mass reaches its lowest point, clearly momentum of the two mass+spring system is not conserved. Including the Earth is not being picky, it's essential.
To avoid that, we can use a very short time interval, making the assumption that the coalescence is achieved quickly.
In between these extremes, we could model the coalescence as a spring of high constant during compression and zero constant in relaxation.
 
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  • #5
kuruman said:
This reasoning is correct. Over the time interval that the masses stick together they are not displaced appreciably in the gravitational field.
topsquark said:
The momentum of the 1.0 kg mass is clearly not conserved. Neither is the momentum of the 2.0 kg mass. But if you take the 1.0 kg mass, the 2.0 kg mass, and the spring (which is ideal, and thus massless) to be your system, then the momentum is conserved, (If you want to be picky, throw the Earth into this and use the Newtonian gravitational potential energy.)

-Dan
haruspex said:
That doesn’t really help. The issue is the time interval over which momentum is to be considered.
If you take the time up until the dropped mass reaches its lowest point, clearly momentum of the two mass+spring system is not conserved. Including the Earth is not being picky, it's essential.
To avoid that, we can use a very short time interval, making the assumption that the coalescence is achieved quickly.
In between these extremes, we could model the coalescence as a spring of high constant during compression and zero constant in relaxation.
Thank you for your replies @kuruman, @topsquark and @haruspex!
 

FAQ: Spring momentum conservation problem

1. What is the principle of momentum conservation in a spring system?

The principle of momentum conservation in a spring system states that the total momentum of the system remains constant if no external forces are acting on it. This means that the sum of the momenta of all objects involved before and after an interaction, such as a collision or compression/expansion of a spring, must be equal.

2. How does Hooke's Law relate to spring momentum conservation problems?

Hooke's Law relates to spring momentum conservation problems by describing the force exerted by a spring as proportional to its displacement from the equilibrium position, given by F = -kx, where k is the spring constant and x is the displacement. This force can affect the momentum of objects attached to the spring, influencing the overall momentum conservation analysis.

3. What are common assumptions made in spring momentum conservation problems?

Common assumptions made in spring momentum conservation problems include: the system is isolated with no external forces acting on it, the spring follows Hooke's Law, the spring is massless or its mass is negligible, and friction or air resistance is ignored. These assumptions simplify the analysis and allow for the application of conservation laws.

4. How do you solve a problem involving a spring and two colliding masses?

To solve a problem involving a spring and two colliding masses, you typically follow these steps: 1) Identify the initial momenta of the masses before the interaction.2) Apply the conservation of momentum principle to find the final velocities of the masses.3) Use Hooke's Law to relate the spring force to the displacement and any potential energy stored in the spring.4) Solve the resulting equations to find the unknown quantities, such as final velocities or spring compression.

5. Can energy conservation be used in conjunction with momentum conservation in spring problems?

Yes, energy conservation can be used in conjunction with momentum conservation in spring problems. While momentum conservation deals with the total momentum of the system, energy conservation focuses on the total mechanical energy, including kinetic and potential energy. By combining both principles, you can solve for unknowns more effectively, especially when dealing with elastic collisions and spring compression/expansion.

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